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The maximum peak-to-peak voltage of an `AM` wave is `16 mV` and the minimum peak-to-peak voltage is `4 mV`. The modulation factor is equal toA. `2/3`B. `2/5`C. `1/3`D. `3/5` |
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Answer» Correct Answer - D `E_"max"=16/2`=8 mV `E_"min" = 4/2 =2 ` mV `therefore mu=(E_"max"-E_"min")/(E_"max"+E_"min") =(8-2)/(8+2)=6/10=3/5` |
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